A block of ice was melting at a steady rate. At 10 a.m. the ice weighted 3kg and at 10:05 am, it weighted 2.25kg. What was the melting rate in kg/min?
2006-10-07 13:59:05 · 7 answers · asked by Jackie C 1 in Science & Mathematics ➔ Mathematics
we're looking for the mass change per minute.
In 5 minutes, the mass change was 3kg - 2.25kg = 0.75kg.
So that was 0.75kg per 5 mins.
So in one minute, it would be a fifth of that, or 0.15kg/min.
Reading James L's answer below, he's assuming an exponential loss in mass, instead of a linear one. The wording in your question was 'steady rate', so I think we're OK with the 0.15kg/min answers.
2006-10-07 14:02:59 · answer #1 · answered by georgex 2 · 1⤊ 1⤋
3kg-2.25kg=.75kg melted off the ice in 5 minutes
so .75kg divided by 5 minutes = .15kg per minute
2006-10-07 14:06:09 · answer #2 · answered by Tommiecat 7 · 0⤊ 0⤋
In five minutes, the ice lost .75 kg.
You want to know how many kg in 1 minute.
In other words, if it takes 5 minutes to lose .75, then is takes 1 minute to lose ____.
You would divide by 5 to get to 1 minute.
So you divide by five to get the right about of kg.
I will leave the division to you.
Good Luck!
2006-10-07 14:05:02 · answer #3 · answered by TripleFull 3 · 0⤊ 0⤋
The mass at time t is M(t)=M0*e^(-kt), where M0=3 kg, and t is the time since 10am, in minutes.
To get k, use M(5)=3*e^(-5k)=2.25, and solve for k. To do this, take the natural log of both sides of e^(-5k)=2.25/3 and get
-5k = ln(3/4),
so k = 0.0575.
Therefore, at 10:05am, the cooling rate is equal to k*M(5) = 0.1295 kg/min.
2006-10-07 14:05:36 · answer #4 · answered by James L 5 · 0⤊ 1⤋
Since (3-2.25)kg = 0.75kg melted in 5 minutes,
the melting rate was 0.75kg / 5 min = 0.15 kg/min.
2006-10-07 14:03:58 · answer #5 · answered by David Y 5 · 0⤊ 0⤋
3 - 2.25 = .75 .75 divided by 5 = .15
2006-10-07 14:15:05 · answer #6 · answered by helper 1 · 0⤊ 0⤋
Rate = 3 - 2.25/ 5 = 0.75/ 5 = 0.15kg/min
2006-10-07 14:33:46 · answer #7 · answered by Hardrock 6 · 0⤊ 0⤋